Show that the dimension of the set of all linear transformations from $V$ to $W$ has dimension equal to $mn$.

Question

Let $V, W$ be vector spaces. The set $L(V,W)$ is the set of all linear transformations $T: V \to W$, with the operations:

$(T_1 + T_2)(v) = T_1(v)+T_2(v)$

$(\alpha T)(v) = \alpha T(v)$

for $v\in V$.

Prove that the dimension of the set $L(V,W)$ is $\mathrm{dim}(L(V,W)) = \mathrm{dim}(V) \cdot \mathrm{dim}(W)$

I know I have to (or can) use the bases of the vector spaces to prove this but I cannot come up with a solution.

Answer 1

A longer hint:

Pick a basis $\{e_i\}$ of V and a basis $\{f_j\}$ of W. Consider the set of linear transformations $S=\{T_{mn}\}$ where we define $T_{mn}$ as $$T_{mn}(\sum a_ie_i) =a_mf_n$$ In other words, $T_{mn}$ maps $e_m$ to $f_n$ and maps the other basis vectors of V to 0 (in W), and then we extend this over all of V while keeping $T_{mn}$ linear.

Can you show that the transformations in S are independent i.e. you cannot make one of them by adding together multiples of the others ?

Can you see how to add together multiples of transformations in S to create a transformation that maps $e_m$ in V to any given vector w in W ?

Now note that a linear transformation in T is defined by its action on each of the basis vectors in V. Can you see how to add together multiples of transformations in S to create any given linear transformation in T ?

Once you have shown that the transformations in S are independent and that they cover all of T, then you know that S is a basis of T. Count the size of S to find the dimension of T

Answer 2

Hint:

By definition of a basis, if $\mathcal B$ is a basis of $V$, and we set $m=\dim V$, $$\mathcal L(V,W)\simeq W^{\mathcal B}\simeq W^m,\enspace\text{so}\quad \dim\mathcal L(V,W)=\dim W^m=m\dim W.$$

Answer 3

This explanation helps if you know that two finite-dimensional vector spaces over the same field, $F$, are isomorphic if and only if they have equal dimensions.

If you're using Sheldon Axler's Linear Algebra Done Right (3rd. Edition), you might have seen $\mathcal{L}(V,W)$ as a domain in the context of using the linear map $\mathcal{M}$, which maps $\mathcal{L}(V,W)$ to $F^{mn}$, where $F^{mn}$ is the set of all m-by-n matrices, where each entry is in $F$.

Formally:

$\mathcal{M} : \mathcal{L}(V,W) \rightarrow F^{mn}$

So using the Fundamental Theorem of Linear Maps, we know that:

dim $\mathcal{L}(V,W) =$ dim null $\mathcal{M}$ $+$ dim range $\mathcal{M}$

So our goal now is to prove that: dim $\mathcal{L}(V,W) =$ (dim $V$)(dim $W$)

Let's start with a basis $v_1, ..., v_n$ for $V$ and a basis $w_1, ..., w_m$ for a basis of W. This means that dim $V = n$ and dim $W = m$. And this also makes sense because $\mathcal{M}(T)$, where $T \in \mathcal{L}(V,W)$, represents an m-by-n matrix that holds the coefficients for the basis of $W$ to make $Tv_k$:

$Tv_k = A_{1,k}w_1 + A_{2,k}w_2 + ... + A_{m,k}w_m$, where $A_{j,k}$ are entires of the k-th column of $\mathcal{M}(T)$ for $1 \leq j \leq m$

If you need help convincing yourself that dim $F^{mn} = mn =$ (dim $V$)(dim $W$), just imagine the basis of $F^{mn}$ to be a list of m-by-n matrices where each matrix has entries of all $0$'s except one where there is a $1$:

$\begin{equation} \begin{bmatrix} 1 & 0 & \ldots & 0\\ 0 & 0 & \ldots & 0\\ \vdots & \ldots & \ldots & 0\\ 0 & \ldots & \ldots & 0\\ \end{bmatrix}, \ \begin{bmatrix} 0 & 1 & \ldots & 0\\ 0 & 0 & \ldots & 0\\ \vdots & \ldots & \ldots & 0\\ 0 & \ldots & \ldots & 0\\ \end{bmatrix}, \ \ldots , \ \ \begin{bmatrix} 0 & 0 & \ldots & 0\\ 0 & 0 & \ldots & 0\\ \vdots & \ldots & \ldots & 0\\ 0 & \ldots & \ldots & 1\\ \end{bmatrix} \end{equation}$

We see that there are a total of $mn$ matrices in the basis of $F^{mn}$.

Now, in order for dim $\mathcal{L}(V,W) = mn =$ (dim $V$)(dim $W$), we need to show that dim $\mathcal{L}(V,W) = $ dim $F^{mn}$. So that means that we need to show that dim null $\mathcal{M} = 0$ and dim range $\mathcal{M} = $ dim $F^{mn}$ (in other words, show that $\mathcal{M}$ is injective and surjective).

To show that $\mathcal{M}$ is injective, we need to show that $\mathcal{M}(T) = 0$, where $T = 0$. In order to do so, let's suppose that $\mathcal{M}(T) = 0$, where $0$ is the m-by-n matrix that has all entries as $0$. What this means is that: $Tv_k = 0$ because:

$Tv_k = 0w_1 + 0w_2 + ... + 0w_m$

Since $v_k$ is an element from the basis of $V$, $v_k \neq{0}$, therefore $T = 0$. Hence, null $\mathcal{M} = \{0\}$ and thus dim null $\mathcal{M} = 0$.

So what we have now is dim $\mathcal{L}(V,W) = $ dim range $\mathcal{M}$, where we need to show that range $\mathcal{M} = F^{mn}$.

Let's suppose that $A \in F^{mn}$. Remember that:

$Tv_k = A_{1,k}w_1 + A_{2,k}w_2 + ... + A_{m,k}w_m$, where $A_{j,k}$ are entires of the k-th column of $\mathcal{M}(T)$ for $1 \leq j \leq m$

So, we see that $\mathcal{M}(T) = A$. By using the element-method, we see that range $\mathcal{M} \subseteq F^{mn}$ and $F^{mn} \subseteq$ range $\mathcal{M}$, thus range $\mathcal{M} = F^{mn}$.

Therefore, dim $\mathcal{L}(V,W) = $ dim $F^{mn} = mn = $ (dim $V$)(dim $W$).

As we see, we showed that because these two vector spaces are isomorphic, they possess equal dimensions.